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Chapter 7 · Kerala SSLC Class 10 Maths

Coordinates

Two ways to learn — a crisp formula reference, or a story that builds coordinate geometry from one simple rule about axis-parallel lines.

Rectangle verticesDistance formulaDistance from originCollinearityFourth-vertex trick
Kerala SSLCClass 10MathematicsChapter 79 min crisp · 14 min story

Four scenes from a rectangle's missing corners to the fourth-vertex trick.

How this works

Four scenes build coordinate geometry from one simple rule about axis-parallel lines, all the way to an elegant trick for a point trapped inside a rectangle. Read the story, then tap "The Maths" for the formal result.

1

Finding the Missing Corners

Every point on a flat surface can be located with two numbers — its coordinates — measured from two perpendicular reference lines: the horizontal x-axis and the vertical y-axis, crossing at a point called the origin, O.

Here's a rectangle drawn with its sides running exactly parallel to these axes. Two of its opposite corners are given: (2, −1) and (7, 5). Can you find the other two corners without drawing anything?

The rule

Moving parallel to the x-axis never changes the y-coordinate. Moving parallel to the y-axis never changes the x-coordinate.

So the corner sharing the left side with (2, −1) must also have x = 2 — and since it's level with (7, 5), its y-coordinate must be 5. That gives (2, 5). The remaining corner, by the same logic, is (7, −1).

Given opposite corners: (2, −1) and (7, 5)

Missing corners: (2, 5) and (7, −1)

2

The Diagonal Is a Hypotenuse

Once you can find a rectangle's four corners, its side lengths follow just as easily — and that unlocks something more useful: the distance between the two opposite corners themselves.

Take the same rectangle, opposite corners (2, −1) and (7, 5). Its horizontal side runs from x = 2 to x = 7 — length 5. Its vertical side runs from y = −1 to y = 5 — length 6.

horizontal side = |7 − 2| = 5

vertical side = |5 − (−1)| = 6

The line joining (2, −1) and (7, 5) is exactly the diagonal of this rectangle — and a rectangle's diagonal is the hypotenuse of a right triangle formed by two adjacent sides.

diagonal² = 5² + 6² = 25 + 36 = 61

diagonal = √61

One special case is worth naming on its own: the distance from any point (x, y) to the origin itself.

Treating O(0,0) and (x,y) as opposite rectangle corners:

side lengths are |x| and |y|

distance to origin = √(x² + y²)

3

Three Points, One Line?

A different kind of question: are the points A(−1, 2), B(3, 5), and C(9, −3) all on the same straight line?

The idea

If three points sit on the same line, one of them sits exactly between the other two — meaning the longest of the three distances between them equals the sum of the shorter two, just like points on a number line.

AB = √((−1−3)² + (2−5)²) = √(16+9) = √25 = 5

BC = √((3−9)² + (5−(−3))²) = √(36+64) = √100 = 10

AC = √((−1−9)² + (2−(−3))²) = √(100+25) = √125 ≈ 11.18

The largest distance here is AC ≈ 11.18. Is that equal to AB + BC = 5 + 10 = 15?

No — 11.18 ≠ 15, so A, B and C are NOT on the same line.

4

The Point Inside the Rectangle

The chapter's trickiest and most elegant problem: a point P sits inside a rectangle. Its distances to three of the four corners are known — 3 cm, 4 cm, and 5 cm. What is its distance to the fourth corner?

Set up axes along two sides meeting at corner O, so O = (0,0), the corner along the x-axis is A = (a,0), the corner along the y-axis is C = (0,b), and the far corner is B = (a,b). Let P = (x,y).

PC² = x² + (y−b)² = 9

PO² = x² + y² = 16

PA² = (x−a)² + y² = 25

We want PB² = (x−a)² + (y−b)². Add the PC² and PA² equations together:

PC² + PA² = [x² + (y−b)²] + [(x−a)² + y²]

= [x² + y²] + [(x−a)² + (y−b)²]

= PO² + PB²

So PC² + PA² = PO² + PB² — the two DIAGONAL pairs of corners give equal sums of squared distances. Substituting the known values:

9 + 25 = 16 + PB²

34 = 16 + PB²

PB² = 18

PB = √18 = 3√2 cm

The Big Picture

Everything in this chapter grows out of one rule: moving along an axis-parallel line keeps one coordinate fixed. Master these five moves and Chapter 7 is done.

Q

Missing two corners of a rectangle?

(x₁,y₂) and (x₂,y₁)

Q

Distance between two points?

d = √((x₁−x₂)² + (y₁−y₂)²)

Q

Distance from the origin?

d = √(x² + y²)

Q

Are three points on one line?

largest distance = sum of other two

Q

Distance to a rectangle's 4th corner?

PW²+PY² = PX²+PZ² (diagonal pairs)

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